91Ó°ÊÓ

The natural logarithm $$\ln x$$ is undefined for x ≤ 0. Given that the function is $$g(x)=2-\ln x^{2}$$, the natural logarithm $$\ln x^{2}$$ is undefined for x = 0.

For the undefined point x = 0, we need to check if the function approaches infinity (positive or negative) as x approaches this value. Let's analyze the limits from the left and right of this point: $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (2 - \ln x^2)$$ $$\lim_{x \to 0^+} g(x) = 2 - \lim_{x \to 0^+} \ln x^2$$ As x approaches 0 from the right, $$x^2$$ approaches 0, so the natural logarithm approaches negative infinity: $$\lim_{x \to 0^+} \ln x^2 = -\infty$$ Thus, the limit becomes: $$\lim_{x \to 0^+} g(x) = 2 - (-\infty) = \infty$$ Now let's analyze the limit from the left of the point x=0: $$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (2 - \ln x^2)$$ Since x^2 is always positive, even if x is negative, the limit from the left will be the same as from the right. So: $$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = \infty$$

Since the function approaches infinity (positive) as x approaches 0 from both the left and right, there is a vertical asymptote at x=0.